Assume that $C$ is a negatively oriented, piecewise smooth, simple, closed curve. Let $R$ be the region enclosed by $C$. Use the circulation form of Green's theorem to rewrite $ \oint_C (3x^2 + x^3) \, dx + \cos(xy^2) \, dy$ as a double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \iint_R -6x + 3x^2 + 2xy\sin(xy^2) \, dA$ (Choice B) B $ \iint_R 6x - 3x^2 + 2xy\sin(xy^2) \, dA$ (Choice C) C $ \iint_R -2xy\sin(xy^2) \, dA$ (Choice D) D $ \iint_R y^2\sin(xy^2) \, dA$ (Choice E) E Green's theorem is not necessarily applicable.
Explanation: Assume we have a two-dimensional vector field $F(x, y) = P(x, y) \hat{\imath} + Q(x, y) \hat{\jmath}$ and a piecewise smooth, simple, closed curve $C$. Let $R$ be the region enclosed by $C$. Then the circulation form of Green's theorem states that we have the equality below: $ \oint_C P \, dx + Q \, dy = \iint_R \left( \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} \right) dA$ Our first step should be to confirm that the given curve is compatible with using Green's theorem. The curve $C$ does satisfy all the conditions of Green's theorem. It is also negatively oriented, which means we'll need to take the negative of our result at the end. The next step is to identify the components $P$ and $Q$ of the vector field $F$ over which we're integrating. We're simplifying from a line integral to a double integral, so we just need to match up the $dx$ term to $P$ and the $dy$ term to $Q$ : $\begin{aligned} & \oint_C (3x^2 + x^3) \, dx + \cos(xy^2) \, dy \\ \\ &P(x, y) = 3x^2 + x^3 \\ \\ &Q(x, y) = \cos(xy^2) \end{aligned}$ Our final step is to find $\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}$. $\begin{aligned} &Q_x = -y^2\sin(xy^2) \\ \\ &P_y = 0 \\ \\ &\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} = -y^2\sin(xy^2) \end{aligned}$ We need to take the negative of our result because the curve $C$ is negatively oriented. Putting everything together, we get this double integral: $ \iint_R y^2\sin(xy^2) \, dA$